help me please.
Write a function: User enters natural numbers A and B (A <B). The function prints all prime numbers separated by a space on the segment [A, B]
This is my code but it doesn't work, wrong output, I don't understand why?
def prime_numbers(a, b):
for i in range(a, b + 1):
dividers = 0
for j in range (2, i):
if i % j == 0:
dividers += 1
if dividers == 1:
return i
a = int (input ("Enter first number: ")
b = int (input ("Enter second number: ")
print (prime_numbers (a, b))
When this code is e.g. ran with a=5 and b=11 I expect the output to be 5 7 11. Instead the output is 6.
Problems with Logic
The function you wrote is meant to calculate how many dividers are in the numbers from a-b, which is different than what the prompt is asking.
You're on the right track with the loops and the % (mod) if statement, but you're off by a bit. In order to calculate the number of prime numbers, there are through a-b, you're going to have to do something different because right now I think your code is calculating the number of divisors there are for numbers a-b.
Problems with Code
When you use the keyword return, you return a value for the entire function and considering you're calculating for multiple prime numbers, I don't think that'd work out well.
Also, I'd be sure to watch out for proper tabbing, your tabbing is inconsistent, and sometimes you've used spaces, sometimes you've used tabs. Python doesn't use brackets (languages such as C, C++, Java, JavaScript aren't reliant on tabs) and is reliant on consistent spacing/tabbing.
Don't forget to close the brackets for the int() function, which is meant to convert a string to an integer.
I think what you meant to do is this:
def prime_numbers(a, b):
primes = [] # Make a list of primes
for i in range(a, b + 1): # Iterate from a-b
if is_prime(i): # Check if the number is a prime or not
primes.append(i) # If it is a prime, then add it to the list
return primes # Return the prime list
def is_prime(v): # Function to check for prime numbers
for x in range(2, v): # Iterate from 2 to the input value
if v % x == 0: # Check if v is divisible by x
return False # If it is divisible by x, return False becaues it's not a prime number
return True # Haven't found any divisors for the number, so return True.
a = int (input ("Enter first number: "))
b = int (input ("Enter second number: "))
print (prime_numbers (a, b))
I'd suggest you use websites such as YouTube and W3Schools to get a better understanding of how to code and remember, GOOGLE IS YOUR FRIEND.
Edit: My apologies, the tabbing issue was caused due to my IDE (Sublime), but it's still a good thing to keep in mind when working with Python.
Related
I am trying to make a simple function to sum the prime numbers of a given input. I am wondering why '7' isn't coming up in my appended list of prime numberS:
def sum_primes(n):
empty = []
for i in range(2,n):
if n % i == 0:
empty.append(i)
print(empty)
sum_primes(10)
Your method for determining if numbers are prime is a bit off. Your function seems to determine if the input num n is prime but not sum all prime numbers to it.
You could make an isprime function and then loop over that for all numbers less than n to find all primes.
Actually your program have logical error. First you have to understand the prime number logic.
Step 1: We must iterate through each number up to the specified number in order to get the sum of prime numbers up to N.
Step 2: Next, we determine whether the given integer is a prime or not. If it is a prime number, we can add it and keep it in a temporary variable.
Step 3: Now that the outer loop has finished, we can print the temporary variable to obtain the total of primes.
sum = 0
for number in range(2, Last_number + 1):
i = 2
for i in range(2, number):
if (int(number % i) == 0):
i = number
break;
if i is not number:
sum = sum + number
print(sum)
def sum_primes(y):
sum = 0
# for each number in the range to the given number
for i in range(2, y):
# check each number in the sub range
for j in range(2, int(i/2)+1):
# and if its divisble by any number between
# 2 and i/2
if i % j == 0:
# then it is not a prime number
break
else:
# print current prime
print(i)
# add prime to grand total
sum += i
return sum
print(sum_primes(10))
I think you have some problems with the way you are checking if your number is prime or not. A much simpler way would be to use a library that does this for you. It takes less time, is usually a better implementation than you or I could come up with, and it takes fewer lines, therefore, looking cleaner.
I would recommend primePy. Its whole point is to work with prime numbers.
I don't have my computer with me atm so I can't make any sample code for you and test it. But something like this should work.
from primePy import primes
prime_list = primes.upto(n)
print(prime_list)
I believe that function returns a list of all prime numbers up to n. And if you wanted to get the sum of all of those numbers, you can go about whatever method you want to do that.
Afterward, slap it into a def and you should be good to go.
If it doesn't work let me know, and I will try to test and fix it when I am in front of a computer next.
The output shows a different result. Yes, the factorials of those numbers are right but the numbers outputted aren't right.
Here's the code:
input:
n = int(input("Enter a number: "))
s = 0
fact = 1
a = 1
for i in range(len(str(n))):
r = n % 10
s += r
n //= 10
while a <= s:
fact *= a
a += 1
print('The factorial of', s, 'is', fact)
Output:
Enter a number: 123
The factorial of 3 is 6
The factorial of 5 is 120
The factorial of 6 is 720
You're confusing yourself by doing it all in one logic block. The logic for finding a factorial is easy, as is the logic for parsing through strings character by character. However, it is easy to get lost in trying to keep the program "simple," as you have.
Programming is taking your problem, designing a solution, breaking that solution down into as many simple, repeatable individual logic steps as possible, and then telling the computer how to do every simple step you need, and what order they need to be done in to accomplish your goal.
Your program has 3 functions.
The first is taking in input data.
input("Give number. Now.")
The second is finding individual numbers in that input.
for character in input("Give number. Now."):
try:
int(character)
except:
pass
The third is calculating factorials for the number from step 2. I won't give an example of this.
Here is a working program, that is, in my opinion, much more readable and easier to look at than yours and others here. Edit: it also prevents a non numerical character from halting execution, as well as using only basic Python logic.
def factorialize(int_in):
int_out = int_in
int_multiplier = int_in - 1
while int_multiplier >= 1:
int_out = int_out * int_multiplier
int_multiplier -= 1
return int_out
def factorialize_multinumber_string(str_in):
for value in str_in:
print(value)
try:
print("The factorial of {} is {}".format(value, factorialize(int(value))))
except:
pass
factorialize_multinumber_string(input("Please enter a series of single digits."))
You can use map function to get every single digit from number:
n = int(input("Enter a number: "))
digits = map(int, str(n))
for i in digits:
fact = 1
a = 1
while a <= i:
fact *= a
a += 1
print('The factorial of', i, 'is', fact)
Ok, apart from the fact that you print the wrong variable, there's a bigger error. You are assuming that your digits are ever increasing, like in 123. Try your code with 321... (this is true of Karol's answer as well). And you need to handle digit zero, too
What you need is to restart the calculation of the factorial from scratch for every digit. For example:
n = '2063'
for ch in reversed(n):
x = int(ch)
if x == 0:
print(f'fact of {x} is 1')
else:
fact = 1
for k in range(2,x+1):
fact *= k
print(f'fact of {x} is {fact}')
I'm working on a problem that involves putting in an input, integer n, that when doing so will print off the following 4 "multiples" of the integer. I need to do this for 3 integers, n = 5, n = 0, n = 3.
Original Question:
Implement a program that requests a positive
integer n from the user and prints the first four multiples of n: Test
your module for n = 5; n = 0 and n = 3.
The output of the code should look like:
>>>
Enter n: 5
5
10
15
20
So, what I've come up with so far is this
n = (input("Enter n:"))
This allows me to input an integer value.
Next using print(n), this will print the value I input (Ex. number 5), but I'm not sure how to print off multiples of it after. I realize it's a loop question, most likely involving if or in, but I'm not sure where to go after this.
You've pretty much figured out the question on your own. The correct code is:
n = int(input("Enter n:"))
for i in range(4):
print(n*(i+1))
So, what this for loop does for you is repeat your print statement 4 times, where you give i the values of the expression range(4).
If you just print(range(4)), you'll see that it evaluates to [0,1,2,3]. That's why I had to add 1 to it each time.
The int() function call is needed because input() returns a string, not a number. So if we want the mathematical operators to do what we expect, we need to first convert it to a number (in this case, an integer).
This is the general logic:
n = (input("enter n:"))
for(int i = 1; i <= 4; i++){
print(int(float((n))*i);
}
if you want the list to start with 0 you can do this, it has an error but it can be fixed...
number = int(input("Give a number:"))
for multiples in range(10):
getal1 = number * multiples
print("\t The", str(multiples + 1) + "e multiple of," + number, "is", str(getal1) + ".")
I am making a quick program that will ask the user for a number and then output all the prime numbers up to that number:
n=int(input("Enter a number: "))
a=2
if n<=1:
n=int(input("Enter another number: "))
while a<n:
for i in range(2,n):
if a%i==0:
break
else:
print (a)
break
a=a+1
The problem I am having is that it won't print out 2. For example, if I use 20 as my n value, it will print:
Enter a number: 20
3
5
7
9
11
13
15
17
19
You have two mistakes:
The else is incorrectly indented, so it sits with the if not the for (you want to print if all values below a aren't factors of a, not on the first one that isn't); and
Your inner range goes up to n, not a (so always includes a, and a % a == 0).
This will work:
for a in range(2, n):
for i in range(2, a):
if a % i == 0:
break
else:
print(a)
Note I have used a for loop to replace the outer while, which is generally better when you already know where to stop. You can make the code more efficient by checking up to the square root of a, and leaving out even numbers other than 2, but I will leave those optimisations to you.
You can start the loop from 1.
Your loop will start from 1, hence you will get 2 in output.
This question already has answers here:
Closed 11 years ago.
Possible Duplicate:
Simple Prime Generator in Python
First I will prompt user to input any number. Then my code will check whether does the number input by the user is a Prime number not.
Here is my codes:
num = int(raw_input("Input any of the number you like:"))
for x in range(2, int(num**0.5)+1):
if num % x == 0:
print "It is not a prime number"
else:
print "It is a prime number"
But question is I cant seem to get the output for 2 and 3. And when I randomly input any numbers like 134245, the system will output alot of sentences. And I do not know why?
Appreciate any kind souls to help me up :)
import urllib
tmpl = 'http://www.wolframalpha.com/input/?i=is+%d+a+prime+number'
def is_prime(n):
return ('is a prime number' in urllib.urlopen(tmpl % (n,)).read())
you should stop once num % x == 0 is true (no need for further testing) and print 'it is a prime number' only if the loop completed without anything printed before.
A number is prime if it only divides by 1 and itself. A pseudocode follows:
boolean prime = true;
for (int i = 2; i * i <= num; i++)
if (num % i == 0) {
prime = false;
break;
}
if (prime)
println("It is prime!");
else
println("It is not prime!");
Look at your code like this:
num = ...
for x in range(2, int(num**0.5)+1):
print something
The body of the loop is executed at every iteration. That means you're printing something at every iteration, i.e. for each x that you check to see if it's a factor of num, you print. That's not what you should be doing; in order to determine whether a number is prime, you check all possible factors first, then print your result. So you shouldn't be printing anything until after the loop.
But question is I cant seem to get the output for 2 and 3.
You're looping from 2 to ceil(sqrt(n)). For 2 and 3, this is an empty range, hence no iteration happens. Either special-case it or rewrite the code such that it assumes that n is prime and tries to disprove it in the loop.
the system will output alot of sentences.
You're printing on every iteration. Instead, use a boolean flag (or a premature return, if you factor it out into a function) to determine prime-ness and print once, after the loop, based on that prime.
Your code is not structured well-- the algorithm continues to loop all the way up to the top of your range, even if you already know that the number is composite, and it also prints some result on each iteration when it should not.
You could put the logic into a function and return True or False for prime-ness. Then you could just check the result of the function in your if statement.
def is_prime(num):
for x in range(2, int(num**0.5)+1):
if num % x == 0:
return False
return True
num = int(raw_input("Input any of the number you like:"))
if not is_prime(num):
print "It is not a prime number"
else:
print "It is a prime number"
Here are two Python routines for calculating primes. (Hint: Google for Sieve of Eratosthenese):
def pythonicSieve(maxValue):
"""
A Pythonic Sieve of Eratosthenes - this one seems to run slower than the other.
see http://love-python.blogspot.com/2008/02/find-prime-number-upto-100-nums-range2.html
"""
return [x for x in range(2,maxValue) if not [t for t in range(2,x) if not x%t]]
def sieveOfEratosthenes(maxValue):
"""
see http://hobershort.wordpress.com/2008/04/15/sieve-of-eratosthenes-in-python/
"""
primes = range(2, maxValue+1)
for n in primes:
p = 2
while n*p <= primes[-1]:
if n*p in primes:
primes.remove(n*p)
p += 1
return primes